# How to measure an inductance

Discussion in 'General Electronics Chat' started by Wendy, Apr 24, 2009.

1. ### Wendy Thread Starter Moderator

Mar 24, 2008
20,772
2,540
It has been a while for me, so I'm going through the math again, and putting it up here as a check.

I have some inductance I want to measure. I know the range, but not the value. I'll use a test rig something like this...

The measurements at this point are hypothetical, I'm going through the math. The 100Khz will be from an old function generator, I'll be setting the freq with my counter built into my DVM. The coils resistance is also measure with the DVM.

RL = 1Ω
R1 = 1KΩ
L = ?

Total reactance formula is:

Rt = SQR( (R1+RL)^2 + (2πLf)^2 )

Therefore

L = SQR( (Rt^2 - (R1+RL)^2) / (2πF)^2 ) , Total Reactance (Rt) is the unknown at this point.

Extrapolating VL,

VL = V1 - RL ( (Vt-V1)/R1 ) = 1V - (1Ω (2V/1KΩ)

RL is insignificant in this example, but that won't always be the case.

I'll have to come back on this post to edit it, as I work it out.

• ###### Inductor Measurement.GIF
File size:
2.4 KB
Views:
627
Last edited: Apr 24, 2009

Apr 5, 2008
15,799
2,386
3. ### Wendy Thread Starter Moderator

Mar 24, 2008
20,772
2,540
Doing it my way? I figuring out a practical home lab method.

Apr 5, 2008
15,799
2,386
5. ### Wendy Thread Starter Moderator

Mar 24, 2008
20,772
2,540
Congrats, Bertus!

OK, eliminating RL for the moment, lets go through the math...

V1 = (Vt RL) / ( SQR( R1² + RL²) )
V1² = (Vt² RL²) / (R1² + RL²)
V1² R1² + V1² RL² = Vt² RL²
Vt² RL² = 2 V1² R1²
RL² = (2 V1² R1²) / Vt²
RL = SQR (2) V1 R1 / Vt

If R1 = 1KΩ, Vt = 3V, and V1 = 1V, then RL = 471.4

Rt = SQR( 1KΩ² + 471.4Ω²) = 1105.4Ω

RL is actual XL, so the inductor is 2π(100KHz)L, L = 750.3µH

OK, Lets do the check...

XL = 2πfL = 2π 10Khz 750.3µH = 471.4Ω

Xt = SQR( 1KΩ² + 471.4Ω²) = 1105.5Ω

VL = XL Vt / Xt = 471.4Ω 3V / 1105.5Ω = 1.279V

<dangit, did not work>

I'm missing something obvious. It has to do with how voltage is calculated across the inductor, I'm not interested in the phase angle.

Oh well, time for bed.

Last edited: Apr 24, 2009
6. ### Wendy Thread Starter Moderator

Mar 24, 2008
20,772
2,540
OK, couldn't sleep. Redoing the orignal setup...

Looked up Wikipedia, this article uses the equation (translating the variables to my setup)...

Vo = (XL Vi) / (R1 + XL)
Vr = (R1 Vi) / (R1 + XL)
I = Vi / (R1 + XL)

If Rt = SQR( R1² + XL²), How does this jive with I = Vi / (R1+XL)?

Part of my problem is the voltages are not linear, the voltage across the resistor is not 2V, but a vector derivative, as is the 1V I am hypothetically measuring. So the statement...

Vi = Vr + Vo

is false according to other text books I've been reviewing.

OK, going with the math again...

Vo = (Vi XL) / (R1 + XL)
Vo R1 + Vo XL = Vi XL
Vi XL - Vo XL = Vo R1
XL = (Vo R1) / (Vi - Vo)
L = (Vo R1) / (2πf (Vi - Vo) )

Plugging the numbers in I get 795.8µH

OK, the check...

XL = 2πfL = 500Ω
Xt = SQR( 1000² + 500²) = 1118Ω
I = 3V / 1118Ω = 2.683ma
Vr = 1000Ω 2.683ma = 2.683V
Vo = 500Ω 2.683ma = 1.342V
Vi = SQR( Vr² + Vo²) = 3.00V

DANGIT!

From another angle...

Vr = (R1 Vi) / (R1 + XL) = (1000Ω 3V) / (1000Ω + 500Ω) = 2V

I'm still missing something.

----------------------------

Round 4

OK, lets go with the vector interpretation of the voltages.

Vi = SQR( Vr² + Vo²)
Vi² = Vr² + Vo²
Vr² = Vi² - Vo²
Vr = SQR( Vi² - Vo²) = 2.828V

Vi = SQR( Vr² + Vo²) = 3V

This seems to go somewhere, but I'm going to have to take a break from it.

• ###### Inductor 3rd.GIF
File size:
2.3 KB
Views:
579
Last edited: Apr 24, 2009
7. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Bill,

Have you considered using a squarewave input and calculating the inductance based on the time constant as measured using an o-scope?

hgmjr

8. ### Wendy Thread Starter Moderator

Mar 24, 2008
20,772
2,540
The whole point is simplicity. Once the math is figured out (if I can do it) just a decent meter and sine wave signal source will do it.

All the variables are there, I just need to figure out how to interpret them. My instructor once told me skull sweat is cheaper than hardware, he didn't mention all rust flakes coming off.

Do you think Wikipedia got it wrong on this one? Their math sure was simplier.

9. ### AlexR Well-Known Member

Jan 16, 2008
735
54
The math is simpler.

First get the current through R1; I = (Vt-V1)/R1

Next calculate total impedance; Z= Vt/I

Then find inductance reactance of the inductor; XL = $\sqrt{Z^{2}-R1^{2}}$

If the coil resistance is known and significant it can be added to R1.

If it is not know you will have to make two measurements at different frequencies and set up a pair of simultaneous equations to solve both L and RL.

Once you know XL its a trivial calculation to find L.

Last edited: Apr 24, 2009
10. ### Wendy Thread Starter Moderator

Mar 24, 2008
20,772
2,540

I = (Vi-Vo)/R1 = 2ma

Z = 3V / 0.002A = 1500Ω

XL = SQR( Z² - R1²) = 1118Ω

L = XL / 2πf = 1.779mH

Close enough, I think. I went astray on Round 3, long hours will do that, and rust.

Thanks, now for the experiments.

11. ### Wendy Thread Starter Moderator

Mar 24, 2008
20,772
2,540
OK, a real world problem.

Using a 200µH 5% choke, a 100Khz signal, a 180Ω resistor (measured 177.3Ω), and this DVM, I got these numbers.

Vin = 30.4mv RMS
Vout =15.0mv RMS
I = 86.9µa
Z = 350Ω
XL = 301.8
L = 480µH

A fair call from the part spec. Any ideas?

12. ### Wendy Thread Starter Moderator

Mar 24, 2008
20,772
2,540
Another real world problem.

f = 100Khz
R1 = 177.3Ω
Vin = 31.2mv RMS
Vout = 15.7mv RMS
I = 93.06µa
Z = 335.3Ω
XL = 284.6Ω
L = 453.µH

Obviously the inductance is pretty close.

Last edited: Aug 23, 2009
13. ### KL7AJ AAC Fanatic!

Nov 4, 2008
2,043
293
Another way to do this is to locate a known value of capacitance and look for the resonant frequency. Ths way you don't have to deal with the internal resistance of the inductor, assuming it's relatively high Q.

eric

14. ### Wendy Thread Starter Moderator

Mar 24, 2008
20,772
2,540
In this case the internal resistance was 1.4Ω. I don't think that would account for the amount of error I'm seeing though.

When I get the chance I'll redraw that diagram to change Rt to Z.

15. ### THE_RB AAC Fanatic!

Feb 11, 2008
5,435
1,305
In the interest of simplicity... Many cheap (\$50 ?) digital multimeters these days have capacitance, inductance and even frequency measuring built in.

16. ### Wendy Thread Starter Moderator

Mar 24, 2008
20,772
2,540
Just to restate the methology:

Haven't seen the inductance meters, just the capacitance and freq counter (have one). If I can figure out where I'm going wrong and make it right I'll make an AAC experiment out of it. I figure it is the ratio that is important, so the meter doesn't have to be accurate, just consistant.

My current thought is my function generator has something on the output that is contributing. I'll try a simple transistor buffer, and lowering the freq to see if that makes a difference.

A side thought, it should be able to make a simple 555 Hysteretic Oscillator with a resistor to ground and the coil between pin 3 and the input. That is a different project though.

• ###### temp.GIF
File size:
2.5 KB
Views:
359
Last edited: Aug 24, 2009
17. ### Wendy Thread Starter Moderator

Mar 24, 2008
20,772
2,540
This is my function generator:

Bit of an antique, but it works.

As for the math:

Your equation

L=SQR ((Z²+R1²)/ω²)

My methology (in the illustration):

XL = SQR (Zt²-R1²) .... 2nd equation
L = XL/2πf = XL/ω ...... 3rd equation
L = SQR (Zt²-R1²)/ω = SQR ((Zt²-R1²)/ω²)

We're just merging several step from the methology I showed, so the answer is the same. Just a different breakdown.

Last edited: Aug 24, 2009
18. ### bertus Administrator

Apr 5, 2008
15,799
2,386
Hello Bill,

An other approach is finding the resonance frequqncy of a circuit.

By connecting the known capacitors to the circuits,
you will get the lowest current reading at resonance.

Greetings,
Bertus

File size:
5.8 KB
Views:
339
19. ### davebee Well-Known Member

Oct 22, 2008
539
46
Another thing you could do is to measure the voltage across the R1 resistor directly, then find the resistance that gives the same voltage across R1 when the new resistance replaces the inductor - that resistance will be the impedance of the inductor at 100 kHz.

20. ### Tesla23 Active Member

May 10, 2009
323
67
We have
$
\frac{V_o}{V_i}=\frac{j\omega L}{R_1+j\omega L}
$

you are measuring
$
\alpha = \left|\frac{V_i}{V_o}\right|
$

so

$
\frac{1}{\alpha^2} = \left|\frac{V_o}{V_i}\right|^2= \frac{\omega^2 L^2}{R_1^2+\omega^2 L^2}
$

which simplifies to

$
L = \frac{R_1}{\omega\sqrt{\alpha^2-1}}
$

for your example where R1=180, Vi=30.4mV, Vo=15mV, and freq = 100kHz you get L=163uH.